If $X$ and $Y$ are two CW complexes such that $\Sigma X$ and $\Sigma Y$ are homotopy equivalent, are the spaces $X$ and $Y$ homotopy equivalent? I have seen a counterexample when $\pi_1(X)$ is not zero, so I am really looking for counterexamples where $X$ and $Y$ are $1$ connected.
If the suspensions of two simply connected spaces are homotopy equivalent, are the spaces homotopy equivalent?
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algebraic-topology
homotopy-theory
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1Can you include the counter example you mention in your question? Or post it as an answer to your own question? – 2017-02-18
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0@JoshuaRuiter Here is the link to the [counterexample](http://math.stackexchange.com/questions/11268/is-a-map-a-homotopy-equivalence-if-its-suspension-is-so?rq=1) I have not looked at it closely but I think he gives a counterexample to this. – 2017-02-18
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The double suspension theorem states that if $M$ is an integral homology $n$-sphere, then $\Sigma\Sigma M$ is homeomorphic to $S^{n+2}$.
If the suspension of a manifold $N$ is homotopy equivalent to a manifold, then $N$ is homotopy equivalent to a sphere (see this question). Suppose now that $\pi_1(M) \neq 0$, then $M$ is an integral homology sphere which is not homotopy equivalent to a sphere, and therefore $\Sigma M$ is not homotopy equivalent to a manifold. By the Seifert-van Kampen Theorem, $\Sigma M$ is simply connected.
Therefore $X = \Sigma M$ and $Y = S^{n+1}$ are two non-homotopy equivalent simply connected CW complexes for which $\Sigma X$ and $\Sigma Y$ are homotopy equivalent.
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0Why can't $\Sigma M$ be homotopy equivalent to $S^4$? (This is probably well known, but I don't know it.) – 2017-02-18
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1@JasonDeVito: If the suspension of a manifold $N$ is homotopy equivalent to a manifold, then $N$ is homotopy equivalent to a sphere. Therefore, if $\Sigma M$ is homotopy equivalent to $S^4$, $M$ is homotopy equivalent to $S^3$, but $\pi_1(M) \neq 0$. – 2017-02-18